What are the solutions to the equation x^2-4x+5=0 ?

The solutions to the equation x^2-4x+5=0 are x=2+i,x=2-i

Find the zeros of x^2-4x+5=0

The zeros of x^2-4x+5=0 are x=2+i,x=2-i

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x^{2} - 4 x + 5 = 0

x = 2 + i, x = 2 - i

Solution steps

x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 5 : 2 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 i}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 i}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 i}{2 \cdot 1} : 2 + i

x = \frac{- ( - 4 ) - 2 i}{2 \cdot 1} : 2 - i

The solutions to the quadratic equation are:

x = 2 + i, x = 2 - i

(\frac{( π + i )}{( π - i )})^{100}

(4 x + 1)^{2} = 6

(\frac{1}{2})^{2} + (\frac{1}{3})^{2}

27 s + 9 t - 3

3 x^{2} - 41 x = - 110

What are the solutions to the equation x^2-4x+5=0 ?
The solutions to the equation x^2-4x+5=0 are x=2+i,x=2-i
Find the zeros of x^2-4x+5=0
The zeros of x^2-4x+5=0 are x=2+i,x=2-i